Solve for $x$ : $6x^2 + 12x - 288 = 0$
Answer: Dividing both sides by $6$ gives: $ x^2 + {2}x {-48} = 0 $ The coefficient on the $x$ term is $2$ and the constant term is $-48$ , so we need to find two numbers that add up to $2$ and multiply to $-48$ The two numbers $8$ and $-6$ satisfy both conditions: $ {8} + {-6} = {2} $ $ {8} \times {-6} = {-48} $ $(x + {8}) (x {-6}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 8) (x -6) = 0$ $x + 8 = 0$ or $x - 6 = 0$ Thus, $x = -8$ and $x = 6$ are the solutions.